Find the $$ z $$-score corresponding to the given area. Round to two decimals, if necessary, The area to the left of $$ z $$ is $$ 60 \% $$. $$ z=\square $$ The area to the right of $$ z $$ is $$ 10 \% $$. The area between $$ -z $$ and $$ z $$ is $$ 97 \% $$. $$ z=\square $$

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Alright, I need to find the $$ z $$-scores corresponding to the given areas. Let's tackle each part one by one.

**First Problem:**
- **Given:** The area to the left of $$ z $$ is $$ 60\% $$.
- **To Find:** The $$ z $$-score $$ z $$.

Since the area to the left of $$ z $$ is $$ 60\% $$, this means that $$ z $$ is the 60th percentile in the standard normal distribution. To find the exact $$ z $$-score, I'll use the standard normal distribution table or a calculator that provides the inverse cumulative distribution function.

Looking up $$ 60\% $$ in the standard normal distribution table:
- The table typically provides the cumulative area from the left up to a certain $$ z $$-value.
- For $$ 60\% $$, the corresponding $$ z $$-score is approximately $$ 0.25 $$.

So, $$ z \approx 0.25 $$.

**Second Problem:**
- **Given:** The area to the right of $$ z $$ is $$ 10\% $$.
- **To Find:** The $$ z $$-score $$ z $$.

The area to the right of $$ z $$ being $$ 10\% $$ implies that the area to the left of $$ z $$ is $$ 90\% $$ (since the total area under the curve is $$ 100\% $$).

Using the standard normal distribution table:
- $$ 90\% $$ corresponds to a $$ z $$-score of approximately $$ 1.28 $$.

Therefore, $$ z \approx 1.28 $$.

**Third Problem:**
- **Given:** The area between $$ -z $$ and $$ z $$ is $$ 97\% $$.
- **To Find:** The $$ z $$-score $$ z $$.

The area between $$ -z $$ and $$ z $$ represents the central $$ 97\% $$ of the distribution. This means that $$ 97\% $$ of the data lies within $$ z $$ standard deviations from the mean.

To find $$ z $$, I'll use the property that the area between $$ -z $$ and $$ z $$ is $$ 2 \times \text{Area from } 0 \text{ to } z $$.

So, $$ 2 \times \text{Area from } 0 \text{ to } z = 97\% $$.

This simplifies to:
$$ \text{Area from } 0 \text{ to } z = 48.5\% $$

Looking up $$ 48.5\% $$ in the standard normal distribution table:
- The corresponding $$ z $$-score is approximately $$ 2.05 $$.

Therefore, $$ z \approx 2.05 $$.

**Summary of Findings:**
1. For the area to the left of $$ z $$ being $$ 60\% $$, $$ z \approx 0.25 $$.
2. For the area to the right of $$ z $$ being $$ 10\% $$, $$ z \approx 1.28 $$.
3. For the area between $$ -z $$ and $$ z $$ being $$ 97\% $$, $$ z \approx 2.05 $$.

$$ z = 0.25 $$ $$ z = 1.28 $$ $$ z = 2.05 $$

Find the -score corresponding to the given area. Round to two decimals, if necessary,
The area to the left of is .

The area to the right of is .
The area between and is .

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Find the -score corresponding to the given area. Round to two decimals, if necessary, The area to the left of is . The area to the right of is . The area between and is .